Question: $A=\left[\begin{array}{rr}-3 & 3 & -1 & 4 \\2 & 7 & 3 & 11 \\-9 &8 &1 & 6 \\-16 &3 &-3 & 9 \\23 &-16 &1 & 15\end{array}\right]$ $A_{5,3}=$
Solution: Background An $m\times n$ matrix has $m$ rows and $n$ columns. $A=\left[\begin{array}{rr}A_{1,1} & \cdots & A_{1,n} \\\\\vdots \ & \ddots & \vdots \\\\A_{m,1} &\cdots &A_{m,n}\end{array}\right]$ Therefore, the entry $A_{{c},{d}}$ is located on row ${c}$ and column ${d}$. Finding $A_{5,3}$ $A_{{5},{3}}$ is located on row ${5}$ of $A$ : $\left[\begin{array}{rr}-3 & 3 & -1 & 4 \\2 & 7 & 3 & 11 \\-9 &8 &1 & 6 \\-16 &3 &-3 & 9 \\\ {23} & {-16} & {1} & {15}\end{array}\right]$ $A_{{5},{3}}$ is also located on column ${3}$ of $A$. $\left[\begin{array}{rr}-3 & 3 & {-1} & 4 \\2 & 7 & {3} & 11 \\-9 &8 &{1} & 6 \\-16 &3 &{-3} & 9 \\\ {23} & {-16} & {\text1} & {15}\end{array}\right]$ Therefore, $A_{{5},{3}}={1}$. Summary $A_{5,3}=1$